SOLUTION: How do I write the equation of the ellipse {{{9x^2 + 4y^2 - 36x + 24y + 36 = 0}}} in standard form? Thank-you!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do I write the equation of the ellipse {{{9x^2 + 4y^2 - 36x + 24y + 36 = 0}}} in standard form? Thank-you!      Log On


   



Question 269783: How do I write the equation of the ellipse 9x%5E2+%2B+4y%5E2+-+36x+%2B+24y+%2B+36+=+0 in standard form? Thank-you!
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Here is our equation
9x%5E2+%2B+4y%5E2+-+36x+%2B+24y+%2B+36+=+0
step 1 - rearrange so that x's are together and y's are together as
9x%5E2+-+36x+%2B+4y%5E2+%2B+24y+%2B+36+=+0
step 2 - complete the square on both x's and y's to get
9%28x%5E2+-+4x+%2B+______%29+%2B+4%28y%5E2+%2B+6y+%2B+_____%29+=+-36+%2B+_____+%2B+______
The blanks are waiting for numbers . . .
step 3 - take (1/2) middle term and square it. PUt these into the left side of blanks as

we added 4 into the x's, but there is a 9 in front, so we really added 36. add 36 to the right side to keep it balanced.
we added 9 into the y's, but there is a 4 in front, so we really added 36. add another 36 to the right side to keep it balanced. we get

step 4 - rewrite the left side as 2 binomials squared and simplify the right side as
9%28x-2%29%5E2+%2B+4%28y%2B3%29%5E2+=+36
step 5 - divide by 36 to get
%28x-2%29%5E2%2F4+%2B+%28y%2B3%29%5E2%2F9+=+1